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solar
PV仿真程序,仿真效率高
PV simulation program(solar PV simulink)
- 2013-03-22 13:47:09下载
- 积分:1
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FastICA
盲源分离中最为经典的不动点算法——FastICA,该算法简单,对初学者有很大的帮助(Blind Source Separation in the most classical fixed-point algorithm- FastICA, the algorithm is simple, for beginners there is a great help)
- 2009-03-15 13:00:43下载
- 积分:1
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nulbasis
............Basis for the nullspace
- 2010-03-06 17:24:48下载
- 积分:1
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张正友摄像机标定
说明: 经典相机标定法,张正友标定法的MATLAB实现,值得借鉴(The classical camera calibration method, zhang zhengyou calibration method MATLAB implementation, it is worth learning)
- 2019-06-22 14:05:47下载
- 积分:1
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11111
应用简正波理论求解下列声场:如下图所示波导模型,上层为均匀水层,其厚度为100m,声速为1500m/s;下层为硬质均匀海底,海面和海底均平整。点声源位于30m深处,辐射声波频率为25Hz,编程计算10m和90m深度上的传播损失,分析其差别。(The normal wave theory is applied to solve the following sound fields: the waveguide model shown in the following figure shows that the upper layer is a uniform water layer with a thickness of 100m and a sound velocity of 1500m/s; the lower layer is a hard uniform seabed with smooth surface and seabed. The point source is located at the depth of 30m and the frequency of radiated sound wave is 25Hz. The propagation loss at the depth of 10m and 90m is calculated by programming, and the difference is analyzed.)
- 2018-10-27 11:39:53下载
- 积分:1
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point_jiance
对于点检测分析和算例,方便大家学习和使用.(For point detection analysis and examples to facilitate them to learn and use.)
- 2009-10-25 01:14:41下载
- 积分:1
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f
说明: 偏最小二乘法中的交叉有效性计算的matlab程序,怎么计算出来的预测误差平方和PRESS不太对,偏大;()
- 2008-05-10 15:59:11下载
- 积分:1
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MATLAB-from-entry-to-the-master
对于学习matlab的初学者来说是一本很详细的书,手把手教你从零开始学习matlab。对于高级用户可以直接看实践部分。(For beginners to learn Matlab is a very detailed book, taught you how to start from scratch learning Matlab . Advanced users can see the practical part .
)
- 2012-04-18 16:49:02下载
- 积分:1
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add-all-prime-num-smaller-than-n
add all prime num smaller than n and this is a simple code run by windows cmd
- 2015-03-04 14:31:29下载
- 积分:1
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NR_power-Flow
x0 = ones(2,2) Make a starting guess at the solution
options = optimset( Display , off ) Turn off Display
[x,Fval,exitflag] = fsolve(@myfun,x0,options)
The solution is
x =
-0.1291 0.8602
1.2903 1.1612
Fval =
1.0e-009 *
-0.1619 0.0776
0.1161 -0.0469
exitflag =
1
and the residual is close to zero.
sum(sum(Fval.*Fval))
ans =
4.7915e-020
- 2014-02-22 08:37:54下载
- 积分:1